# Homework Help: Energy conservation problem

1. Apr 16, 2013

### Clever_name

1. The problem statement, all variables and given/known data

see link for a picture of the situation, http://i50.tinypic.com/n5g3z5.jpg

2. Relevant equations

(1/2)mv^(2) , (1/2)k*l^(2)

3. The attempt at a solution

Total energy of the system at the given point of time is

mv^(2)/2+kl^(2)/2 = 0.6*100/2+100*0.25/2 = 30+12.5 = 42.5 J

At the maximum and minimum values of l, v = 0

=> kl2/2 = 42.5

=> l = √(85/k)

=> l = ±0.922 m

maximum l = 0.922m and minimum l = -0.922m

Not terribly confident in this approach however, any guidance would be appreciated. Thanks!

2. Apr 16, 2013

### voko

Is angular momentum conserved here?

3. Apr 16, 2013

### Clever_name

yes indeed, so this is a conservation of angular momentum problem.

Last edited: Apr 16, 2013
4. Apr 16, 2013

### voko

Is this enough to get on with the solution?

5. Apr 16, 2013

### Clever_name

I'm now reading up on angular momentum, but please if you have some tips to get the ball rolling that would be greatly appreciated.

6. Apr 16, 2013

### voko

One thing you need to think about is the angle of the velocity vector with the string at the min/max positions. You need the angle to compute the angular momentum.

7. Apr 17, 2013

### Clever_name

Ok so far i have

(1/2)k(xf^(2)-xi^(2)) = mv*l*sin(60) , I have taken xf=0.5m, so now i solve for xi.

((-(0.6)(10)sin(60)*2)/100) +0.5
which gives me xi = 0.44m

am i on the right track with this problem? any guidance would be appreciated, thanks!

8. Apr 17, 2013

### voko

I do not understand what that equation means. The right hand side is the initial angular momentum. But what is on the left hand side? Is that the potential energy of the spring? How could you possibly equate energy and angular momentum?

9. Apr 17, 2013

### Clever_name

ok to start since angular momentum is conserved do i start out by writing the following equation?
m*v1*l1*sin(60) = m*v2*l2sin(a)=m*v3*l3sin(b),
l2 = min and l3 = max

10. Apr 17, 2013

### Clever_name

I'm guessing i need to use r-polar coordinates here also?

11. Apr 17, 2013

### voko

Think about what I wrote in #6.

12. Apr 17, 2013

### Clever_name

ok so m*v1*l1*sin(60)-m*v3*l3sin(a) =0

divide through by m*v3*l3,
i get (mv1l1)/(mv3l3)(sin(60)-sin(a)) therefore 'a' must be equal to 60 at max L?

13. Apr 17, 2013

### voko

I do not see why that would be true.

14. Apr 17, 2013

### Clever_name

no i haven't

15. Apr 17, 2013

### voko

You mentioned polar coordinates earlier. Do you know what the polar-coordinates representation of a velocity vector looks like?

16. Apr 17, 2013

### Clever_name

yes indeed, you have a velocity in the theta direction and a velocity in the radial direction where the square root of the two terms squared are then the total magnitude of the velocity

17. Apr 17, 2013

### voko

18. Apr 17, 2013

### Clever_name

the radial velocity will be equal to 0, so v(theta) when l is at a maximum will be the velocity of the mass?

19. Apr 17, 2013

### Clever_name

oh no velocity in the radial direction for both L max/min will be 0

20. Apr 17, 2013

### voko

And if the radial velocity at the max/min radii is zero, what is the overall direction of the velocity? What is its angle with the radius at min/max?

21. Apr 17, 2013

### Clever_name

90 degrees! the direction of the velocity would be perpendicular to the length of the cable L

22. Apr 17, 2013

### voko

Very well. That should simplify the angular momentum equation significantly. Now all you need to do is add an equation for conservation of energy, and solve these two equations.

23. Apr 17, 2013

### Clever_name

so that would be,

(1/2)k(xf^(2)-xi^(2))=(1/2)m(v1^(2)+v2^(2))

24. Apr 17, 2013

### voko

The conservation of energy literally means that the sum of kinetic energy and potential energy in one configuration equals the sum of kinetic and potential energies in another configuration. I do not think that the equation in #23 expresses that.

25. Apr 17, 2013

### Clever_name

ok so the equation should look like this,

(1/2)kxi^(2)+(1/2)mv1^(2)=(1/2)kxf^(2)+(1/2)mv2^(2)